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Math Help - tan x>x+x^3/3

  1. #1
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    tan x>x+x^3/3

    Prove that \tan x > x+\frac {x^3}3 if 0<x<\frac {\pi}2.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Prove that \tan x > x+\frac {x^3}3 if 0<x<\frac {\pi}2.
    Using Taylor series with remainder

    \tan x = x + \frac{x^3}{3} + 8 \tan c ( 1 + \tan^2 c) (2 + 3 \tan^2c ) \frac{x^4}{4!}

    where 0<c<\frac {\pi}2. Since

     8 \tan c ( 1 + \tan^2 c) (2 + 3 \tan^2c ) \frac{x^4}{4!} > 0

    for

    where 0<c, x<\frac {\pi}2, the result follows.
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  3. #3
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    Other solutions? (Without Taylor series as we haven't learned it and it is too obvious from it :P)
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  4. #4
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    Hi

    Let f(x) = tan x - x - x^3/3
    Then f'(x) = tanx - x = (tan x -x)(tan x + x)

    For 0 < x < pi/2 tan x > 0 and x > 0 therefore tan x + x > 0
    The sign of f'(x) is the same as g(x) = tan x - x
    g'(x) = tanx > 0
    g is increasing on [0,pi/2[
    g(0) = 0 so g(x) > 0 on [0,pi/2[

    Therefore f'(x) > 0 on [0,pi/2[
    f(0) = 0
    f(x) > 0 on [0,pi/2[
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